Basically the circuit below is what my professor gave us and we need to solve for the the voltage across the 8 ohm resistor, the current through the 2 ohm resistor, power absorbed by the 10 ohm resistor, and the power delivered by the 4 a source. I used regular mesh and I got the wrong values(I checked using KCL and KVL). One thing that I think I got wrong is I assumed that I1 = 3A(feedback). I tried to search the web for a solution and I've seen something called "super mesh". Is this helpful? If yes, can anyone assist me on how to work with it?

Other thing is I want to check if my values are correct using MIT's Circuit Sandbox, however I don't know where should I put the ground so that the circuit will simulate in the way I wanted it to be. Every help is very much appreciated Thanks!

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  • 1
    \$\begingroup\$ Show us what you tried. \$\endgroup\$
    – Dave Tweed
    Aug 31, 2014 at 18:32
  • \$\begingroup\$ I1 = 3A, I2 = 4A, I3 = -4A V8 = (I1 - I3) * 8 = 56V I2ohms = I2 = 4A P1ohm = (I1 - I2) ^ 2 * 1 = 1W basically I'm clueless using mesh in this one \$\endgroup\$ Aug 31, 2014 at 18:42
  • \$\begingroup\$ I would have put the ground at the negative terminal of the 9V dc source. \$\endgroup\$
    – Amit Hasan
    Aug 31, 2014 at 19:12
  • \$\begingroup\$ I1=3 is ok. But you can't just say that I2=4, I3=-4, only that I2-I3=4. \$\endgroup\$ Aug 31, 2014 at 19:36
  • \$\begingroup\$ Full solution is given hope you understand. \$\endgroup\$
    – Amit Hasan
    Sep 1, 2014 at 6:40

1 Answer 1


You don't need meshes. Just calculate everything one source at a time, while keeping the other sources zeroed out (i.e. a voltage source is replaced with a wire, and a current source is replaced with an open (cut) circuit). Then sum the results from each source.

So in this case, the voltage source would see nothing but four resistors in series, and both current sources would see (different sets of) 2 series resistors connected in parallel with 2 other series resistors. Hopefully this helps you see how to "zero" the sources.

I'm not familiar with MIT's Circuit Sandbox, but it shouldn't matter where you put the ground since you just need to check the current flowing in different places - so just pick a spot and call it ground!

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    \$\begingroup\$ This is applying the superposition theorem, in case OP wants to lookup a more detailed explanation. It only applies to linear systems. \$\endgroup\$
    – ACD
    Sep 2, 2014 at 17:58

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